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A 30-g bullet traveling horizontally at 300 m/s strikes a 1.0-kg block which is attached to a horizontal spring with a force constant of 2000 N/m and rests on a frictionless horizontal surface. The spring is on the far side of the block and aligned with the direction of travel of the bullet. The bullet becomes embedded in the block and, as a result of the impact, and the block slides against the spring. How far is the spring compressed before it reverses its direction of travel

Sagot :

Answer:

Explanation:

Using law of conservation of momentum during collision , velocity after collision can be found .

common velocity V = m v / ( m + M )

= .030 x 300 / ( .030 + 1 )

V= 8.7378 m /s

The kinetic energy of the bullet and block will be stored as elastic energy

1/2 ( M + m ) V² = 1/2 k A² where k is force constant and A is maximum compression in the spring .

( M + m ) V² =  k A²

1.030 x 8.7378² = 2000 x A²

A² = .0393

A= .1982 m .

= 19.82 cm

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